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# How many five-digit numbers can be formed using the digits

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Manager
Joined: 30 Jul 2007
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How many five-digit numbers can be formed using the digits [#permalink]

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10 Sep 2008, 18:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

* 15
* 96
* 120
* 181
* 216
Director
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10 Sep 2008, 19:58
1+2+3+4+5 is divisible by 3, we can form 5! numbers from 1...5.

Considering 0 as one of the digits, the remaining 4 digits must be 5,4,2,1 in order to the number be divisible by 3

for this, first digit cant be 0,
so 4*4*3*2*1= 96 possibilities

thus total numbers that can be formed are 5! + 96 = 216 possibilities.

E
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10 Sep 2008, 23:43
alpha_plus_gamma wrote:
1+2+3+4+5 is divisible by 3, we can form 5! numbers from 1...5.

Considering 0 as one of the digits, the remaining 4 digits must be 5,4,2,1 in order to the number be divisible by 3

for this, first digit cant be 0,
so 4*4*3*2*1= 96 possibilities

thus total numbers that can be formed are 5! + 96 = 216 possibilities.

E

Perfect. I missed the possiblity of 0 as one of the digits (excluding the first one) and ended up with the answer as 5!
Re: m04 - digits   [#permalink] 10 Sep 2008, 23:43
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